A Charged Isolated Conductor — Question 9

Question 9

Connected unequal spheres share charge but not equal surface fields

Two conducting spheres of radii R and 3R are joined by a thin wire and separated by a distance much larger than either radius. Their total charge is $+8.00$ nC. Neglect mutual distortion and wire charge. Use the supplied equilibrium relation $Q_1/R_1=Q_2/R_2$.

1. Find the final charges and their uniform surface charge densities in terms of R.

2. Compare the surface fields. If the same field limit applies to both spheres, identify which sphere reaches it first as total charge is increased.

Question 9 - Solution

Connected unequal spheres share charge but not equal surface fields

(a) Method and calculation

$$Q_2=3Q_1,\quad Q_1+Q_2=8.00\,\mathrm{nC}\Rightarrow Q_1=2.00,\ Q_2=6.00\quad(\mathrm{nC}).$$$$\sigma_1=\frac{2.00\,\mathrm{nC}}{4\pi R^2},\quad\sigma_2=\frac{6.00\,\mathrm{nC}}{36\pi R^2}.$$

(b) Interpretation and further calculation

$$E_i=\sigma_i/\epsilon_0,\quad\frac{E_1}{E_2}=\frac{Q_1/R^2}{Q_2/(9R^2)}=3.$$$$E_1=3E_2\Rightarrow\text{the smaller sphere reaches the common limit first}.$$The supplied sharing relation is an additional equilibrium condition; charge conservation and Gauss law alone do not determine the partition.

Physical check

The approximation explicitly excludes the strong surface distortion that would occur if the spheres were close together.